The enthalpy change on freezing from 10°C to –10°C may be expressed as Liquid (10°C) $\xrightarrow{\Delta H_1}$ Liquid (0°C) $\xrightarrow{\Delta H_2}$ Solid (0°C) $\xrightarrow{\Delta H_3}$ Solid (–10°C) $\Delta H_1$ = $n\mathrm{C_p[H_2O_{(l)}]}$ × ΔT = 1 × 75.3 $\mathrm{J\,mol^{-1}\,K^{-1}}$ × (–10) = –753 $\mathrm{J\,mol}^{-1}$ = – 0.753 $\mathrm{kJ\,mol}^{-1}$ $\Delta H_2$ = $n(-\Delta_{\mathrm{fus}}H)$ = –1 × 6.03 = –6.03 $\mathrm{kJ\,mol}^{-1}$ $\Delta H_3$ = $n\mathrm{C_P[H_2O_{(s)}]}$ = –1 × 36.8 × 10 = –368 $\mathrm{J\,mol}^{-1}$ = –0.368 $\mathrm{kJ\,mol}^{-1}$ ΔH = $\Delta H_1 + \Delta H_2 + \Delta H_3$ = – 0.753 – 6.03 – 0.368 $\mathrm{kJ\,mol}^{-1}$ = –7.151 $\mathrm{kJ\,mol}^{-1}$